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Veer Surendra Sai University of Technology, BurlaDepartment of Mathematics
Lecture notes in Mathematics-I (Dr. S. Mohapatra)
1 Limit of function:
Let f be a real-valued function defined for all points in a neighbourhood N of a point c exceptpossibly at the point c itself. Recall that any open set containing the element a is called theneighbourhood of a. In particular N(a, δ) = (a−δ, a+δ), δ > 0 is called the δ-neighbourhoodof a and N∗(a, δ) = N(a, δ)−{a} = (a− δ, a)∪ (a, a+ δ) is called the deleted neighbourhoodof a. We are assuming that f is defined in the deleted neighbourhood of a in the followingdefinition of limit.
Definition 1 The function f : X → R is said to have a limit at x = a (a may or may notbelongs to X) if given ε > 0, there exists δ, depending upon a and ε, and there exists L ∈ Rsuch that
0 < |x− a| < δ ⇒ |f(x)− L| < ε,
that is,x ∈ (a− δ, a) ∪ (a, a+ δ) ⇒ f(x) ∈ (L− ε, L+ ε),
that is,x ∈ N∗(a, δ) ⇒ f(x) ∈ N(L, ε).
When this happens, we say that the limit L of f exists at x = a and write it as
limx→a
f(x) = L, or, f(x) → L as x → a.
This definition is called the ε− δ definition of limit.One may observe that f need not be defined as a; even it is defined, it is not necessary
that f(a) be equal to L.
Definition 2 The function f is said to tend to +∞ as x tends to c (or in symbols, limx→c
f(x) =
+∞) if for each G > 0 (however large), there exists a δ > 0 such that
f(x) > G, whenever |x− c| < δ.
Similarly, the function f is said to tend to −∞ as x tends to c (or in symbols, limx→c
f(x) = −∞)
if for each G > 0 (however large), there exists a δ > 0 such that
f(x) < −G, whenever |x− c| < δ.
Definition 3 The function f is said to tend to a limit l as x tends to ∞ (or in symbols,limx→∞
f(x) = l) if for each ε > 0, there exists a k > 0 such that
|f(x)− l| < ε, whenever x > k.
Definition 4 The function f is said to tend to +∞ as x tends to ∞ (or in symbols,limx→∞
f(x) = ∞) if for each G > 0 (however large), there exists a k > 0 such that
f(x) > G, whenever x > k.
1.1 Left hand and right hand limits
While defining the limit of a function f(x) as x tends to c, we consider the values of f(x)when x is very close to c. he values of x may be greater or less than c. If we restrict x tovalues less than c, then we say that x tends to c from below or from the left and write itsymbolically as x → c − 0 or simply x → c−. he limt of f(x) with this restriction on x, iscalled the left hand limit. Similarly, if x takes only the values greater than c, then x is said totend to c from above or from the right, and is denoted symbolically as x → c+0 or x → c+.The limit of f(x) with this restriction on x, is called the right hand limit.
Definition 5 A function f is said to tend to a limit l as x tends to c from the �left, if foreach ε > 0, there exists a δ > 0 such that
|f(x)− l| < ε, whenever c− δ < x < c.
In symbols, we then writelimx→c−
f(x) = l.
Definition 6 A function f is said to tend to a limit l as x tends to c from the right, if foreach ε > 0, there exists a δ > 0 such that
|f(x)− l| < ε, whenever c < x < c+ δ.
In symbols, we then writelimx→c+
f(x) = l.
Note: We say limx→c
f(x) exists if and only if both the limits (the left hand and the right hand)
exists and are equal.One-sided infinite limit may also be defined in the same way as above.
Example 1 Let
f(x) =
x2 − a2
x− a, x = a,
b, x = a.
Then show that limx→a
f(x) = 2a, by using ε− δ definition.
Sol. This function defined on R. Take ε > 0. The condition that 0 < |x − a| < δ, impliesx = a. So we may write
x2 − a2
x− a=
(x− a)(x+ a)
x− a= x+ a.
Now|f(x)− 2a| = |(x+ a)− 2a| = |x− a|
Since we need |f(x) − 2a| < ε whenever |x − a| < δ, clearly we can choose any δ such that0 < δ ≤ ε. Hence it follows from the definition that
limx→a
x2 − a2
x− a= 2a.
Note that 2a = limx→a
x2 − a2
x− a= f(a) = b, unless b = 2a.
2
Example 2 Show that limx→a
x3 − a3
x− a= 3a2, by using ε− δ definition.
Sol. Let ε > 0. Take x = a. Then f(x) = x3−a3
x−a= x2 + xa+ a2 and hence∣∣∣∣x3 − a3
x− a− 3a2
∣∣∣∣ = |(x2 − a2) + a(x− a)|
≤ |x− a||x+ 2a|≤ |x− a|(|x− a|+ 3|a|)
As we need δ > 0 such that |x− a| < δ, choosing first δ < 1, the right hand side of the aboveinequality is less than or equal to δ(1 + 3|a|).This gives an idea as to what δ can be choosen for a given ε > 0.
Choose δ = min
(1,
ε
1 + 3|a|
)so that δ ≤ ε
1 + 3|a|.
Now|f(x)− 3a2| < δ(1 + 3|a|) ≤ ε.
This proves the result.We observe that the choice of δ depends not only on ε, but also on a. Although the
natural domain of f is R− {a}.
Example 3 Show that limx→2
√4x+ 1 = 3, by using ε− δ definition.
Sol. Let ε > 0 be given. Now
|√4x+ 1− 3| =
∣∣∣∣(√4x+ 1− 3)(√4x+ 1 + 3)√
4x+ 1 + 3
∣∣∣∣=
4|x− 2|√4x+ 1 + 3
Choose δ < 1. So |x− 2| < δ < 1 implies 1 < x < 3. Hence1√
4x+ 1 + 3<
1√5 + 3
So
|√4x+ 1 + 3| ≤ 4δ√
5 + 3.
In order that4δ√5 + 3
< ε, we choose δ < (√5+3)ε4
.
Now finally we choose δ = min(1, (√5+3)ε4
) so that δ ≤ (√5+3)ε4
.
Hence |√4x+ 1− 3| < 4δ√
5+3< ε.
This proves the result.
Example 4 Evaluate limx→0+
1
1 + e−1/x.
Sol. [As x → 0+, we feel that 1/x increases indefinitely, e1/x increases indefinitely. e−1/x
tends to 1; thus the required limit may be 1.]We have to show that for a given ε > 0,∃ a δ > 0 such that∣∣∣∣ 1
1 + e−1/x− 1
∣∣∣∣ < ε, whenever 0 < x < δ.
3
Now
∣∣∣∣ 1
1 + e−1/x− 1
∣∣∣∣ = ∣∣∣∣ −e−1/x
1 + e−1/x
∣∣∣∣ = 1
1 + e1/x< ε, when 1 + e1/x > 1
εor 1
x> log(1
ε− 1)
⇒ 0 < x <1
log(1/ε− 1), for 0 < ε < 1.
Thus choosing δ = 1log(1/ε−1)
, we see that if 0 < ε < 1,
∣∣∣∣ 11+e−1/x − 1
∣∣∣∣ < ε, when 0 < x < δ.
Again when ε ≥ 1,
∣∣∣∣ 11+e−1/x − 1
∣∣∣∣ < ε ⇒ e1/x > 1ε− 1, which is true for all values of x, so that
for any δ > 0 would work.
Thus for any ε > 0 we are able to find a δ > 0 such that
∣∣∣∣ 1
1 + e−1/x−1
∣∣∣∣ < ε, when 0 < x < δ.
∴ limx→0+
1
1 + e−1/x= 1.
Example 5 Prove that limx→0
x sin1
x= 0.
Sol. Now
∣∣∣∣x sin 1
x
∣∣∣∣ = |x|∣∣∣∣ sin 1
x
∣∣∣∣ ≤ |x|
Thus choosing a δ = ε, we see that
∣∣∣∣x sin 1
x
∣∣∣∣ < ε, when 0 < |x| < δ.
⇒ limx→0
x sin1
x= 0.
Example 6 Show that limx→3
1
(x− 3)4= ∞.
Sol. Let G be any positive number, however large.
Now
∣∣∣∣ 1
(x− 3)4
∣∣∣∣ > G, or1
(x− 3)4> G, when (x− 3)4 < 1
Gor when 0 < |x− 3| < 1
G1/4 .
Choosing δ = 1G1/4 , we get the required result.
Example 7 Show that limx→1
21/(x−1) does not exist.
Sol. We first consider the left hand limit. Let ε > 0 be given. Choosen a positive integer msuch that 1/2m < ε.
Take δ = 1m
and let x satisfy 1− δ < x < 1. Now −δ < (x−1) < 0, and so 1x−1
< −1δ< 0.
Thus |21/(x−1) − 0| = 21/(x−1) < 2−1/δ < 2−m < ε and hence limx→1−
21/(x−1) = 0.
Next, consider x to be on the right of 1.Let δ > 0 be arbitrary and choose a positive integer m0 such that 1
m0< δ. Then if
n ≥ m0, 1+1n∈ (1, 1+ δ) and 2
1
1+ 1n−1 = 2n, which is unbounded. Therefore lim
x→121/(x−1) does
not exist.
Example 8 Find the right and the left hand limits of a function defined as follows
f(x) =
|x− 4|x− 4
, x = 4,
0, x = 4.
4
Sol. When x > 4, |x− 4| = x− 4. ∴ limx→4+
f(x) = limx→4+
|x− 4|x− 4
= limx→4+
x− 4
x− 4= lim
x→4+1 = 1.
Again, when x < 4, |x− 4| = −(x− 4). ∴ limx→4−
f(x) = limx→4−
−(x− 4)
x− 4= lim
x→4−(−1) = −1.
so thatlimx→4+
f(x) = limx→4−
f(x).
Hence limx→4
f(x) does not exist.
Example 9 If limx→a
f(x) exists, prove that it must be unique.
Sol. Let if possible, f(x) tend to limits l1 and l2. Hence for any ε > 0 it is possible to choosea δ > 0 such that|f(x)− l1| < ε/2, when 0 < |x− a| < δ.|f(x)− l2| < ε/2, when 0 < |x− a| < δ.Now |l1− l2| = |l1−f(x)+f(x)− l2| ≤ |l1−f(x)|+ |f(x)− l2| < ε, when 0 < |x−a| < δ.i.e., |l1− l2| is less than any positive number ε (however small) and so must be equal to zero.Thus l1 = l2.
Theorem 1 (without proof) If f and g are two real valued functions defined on some neigh-bourhood of c such that lim
x→cf(x) = l and lim
x→cg(x) = m then
(i) Let α ∈ R. We have limx→c
αf(x) = αl.
(ii) limx→c
(f ± g)x = limx→c
f(x)± limx→c
g(x) = l ±m.
(iii) limx→c
(fg)x = limx→c
f(x). limx→c
g(x) = lm.
(iv) limx→c
(f/g)x = limx→c
f(x)/ limx→c
g(x) = l/m, provided m = 0.
Example 10 Evaluate (i) limx→−1
(x+ 2)(3x− 1)
x2 + 3x− 2, (ii) lim
x→0
√4 + x− 2
x, (iii) lim
x→0+
sin x√x.
Sol. (i) limx→−1
(x+ 2)(3x− 1)
x2 + 3x− 2=
limx→−1
(x+ 2). limx→−1
(3x− 1)
limx→−1
x2 + 3x− 2=
1.(−4)
−4= 1.
(ii) limx→0
√4 + x− 2
x= lim
x→0
√4 + x− 2
x.
√4 + x+ 2√4 + x+ 2
= limx→0
1√4 + x+ 2
=1
4.
(iii) limx→0+
sin x√x
=
(limx→0+
sinx
x
).
(limx→0+
√x
)= 1.0 = 0.
Example 11 Evaluate limx→1
x2 − 1
x− 1.
Sol. Let us evaluate the left hand and right hand limits.When x → 1−, put x = 1− h, h > 0.
limx→1−
(x2 − 1
x− 1= lim
h→0+
(1− h)2 − 1
−h= lim
h→0+
−h(2− h)
−h= lim
h→0+(2− h) = 2.
Again when x → 1+, put x = 1 + h, h > 0.
limx→1+
(x2 − 1
x− 1= lim
h→0+
(1 + h)2 − 1
h= lim
h→0+(2 + h) = 2.
So that both, the left hand and the right hand, limits exist and are equal. Hence limit of thegiven function exists and equals to 2.
5
Example 12 Evaluate limx→0
e1/x
e1/x + 1.
Sol. Now when x → 0+, 1/x → ∞, e−1/x → 0 and x → 0−, 1/x → −∞, e1/x → 0.
∴ limx→0+
e1/x
e1/x + 1= lim
x→0+
1
e−1/x + 1= 1.
and limx→0−
e1/x
e1/x + 1=
0
1= 0.
so that the left hand limit not equal to the right hand limit. Hence limx→0
e1/x
e1/x + 1does not
exist.
Example 13 Find limx→0
exsgn(x+ [x]), where the signum function is defined as
sgn(x) =
1, if x > 0,0, if x = 0,−1, if x < 0,
and [x] means the greatest integer ≤ x.
Sol. Now limx→0−
exsgn(x+ [x]) = limh→0+
e0−hsgn(0− h+ [0− h]) = limh→0+
(−e−h) = −1,
limx→0+
exsgn(x+ [x]) = limh→0+
e0+hsgn(0 + h+ [0 + h]) = limh→0+
eh = 1.
∴ limx→0
exsgn(x+ [x]) does not exist.
Assignment 1
1. Prove the following limits by using ε− δ definition:
(i) limx→2
x2 − 4
x2 − 2x= 2,
(ii) limx→6
√x+ 3 = 3,
(iii) limx→a
xn = an,
(iv) limx→a
f(x) = a2, where f : R → R be given by f(x) = x2,
(v) limx→a
f(x) = 1/a, where f : (0,∞) → R be given by f(x) = x−1.
2. Evaluate the following limits (i)-(vi), if they exist:
(i) limx→0
3x+ |x|7x− 5|x|
, (ii) limx→1
1
x− 1
(1
x+ 3− 2
3x+ 5
),
(iii) limx→0
1− 2 cos x+ cos 2x
x2, (iv) lim
x→∞
ex − e−x
ex + e−x,
(v) limx→0
tan x− x
x(1− cos x), (vi) lim
x→1
1 + cosπx
tan2 πx,
(vii) Show that limx→0
xe1/x
1 + e1/x= 0,
(viii) Show that limx→0
e1/x − 1
e1/x + 1does not exist,
(ix) Show that limx→0
e1/x − e−1/x
e1/x + e−1/xdoes not exist,
(x) If limx→c
f(x) = l then show that limx→c
|f(x)| = |l|.
1.2 Limit of a function by sequential approach
Definition 7 Let J ⊂ R be an interval. Let a ∈ J . Let f : J\{a} → R be given. Thenlimx→a
f(x) = l iff for every sequence {xn} with xn ∈ J\{a} with the property that xn → a, we
have f(xn) → l.
6
Theorem 2 A function f tends to finite limit as x tends to c if and only if for every ε > 0∃ a neighbourhood N(c) of c such that |f(xm)− f(xn)| < ε for all xm, xn ∈ N(c);xm, xn = c.
Similarly, a function f tends to a finite limit as x tends to ∞ if and only if for every ε > 0,there exists G > 0 such that |f(xm)− f(xn)| < ε, for all xm, xn > G.
Example 14 Show that limx→0
1
xsin
1
xdoes not exist.
Sol. Let f(x) = 1xsin 1
x. The function f is defined for every non-zero real number.
Now for each natural number n, let xn = 2π(4n+1)
, and so f(xn) = (4n+1)π2
sin(2nπ + π2) =
(4n+1)π2
→ ∞ as n → ∞.∴ lim
n→∞f(xn) = ∞, when {xn} = { 2
(4n+1)π} converges to zero.
Again, by taking xn = 1/nπ, we see that f(xn) = nπ.0 = 0 for every natural number n, andso lim
n→∞f(xn) = ∞, when {xn} = {1/nπ} converges to zero.
Therefore, limx→0
f(x) does not exist.
Example 15 Find limx→−∞
x2sgn(cosx).
Sol. Here, f(x) = x2sgn(cosx). Let xn = −2nπ, so {xn} → −∞, as → ∞.Now f(xn) = (−2nπ)2sgn(cos(−2nπ)) = 4n2π2, and so lim
n→∞f(xn) = ∞, when {xn} =
{−2nπ} → −∞.∴ lim
x→−∞x2sgn(cosx) = ∞.
Again, taking xn = −(2n + 1)π, we see that f(xn) = [−(2n + 1)π]2sgn(cos(−(2n + 1)π)) =−(2n+ 1)2π2 and so lim
x→−∞x2sgn(cosx) = −∞.
Hence limx→−∞
x2sgn(cosx) does not exist.
Assignment 2
(i) The function f : R\{0} → [−1, 1] defined by f(x) = sin(1/x) does not exist at x = 0 byusing sequential approach.(ii) Use sequential limit form to obtain the limit of f(x) = x3 + x2 − 5 at x = a.
Lemma 1 (Sandwich Theorem) Let J ⊂ R. Let f, g, h be defined on J\{a}. Assume that(i) f(x) ≤ h(x) ≤ g(x), for x ∈ J, x = a.(ii) lim
x→af(x) = l = lim
x→ag(x).
Then limx→a
h(x) = l.
2 Continuity
Let f be a real-valued function defined on an interval J ⊂ R. We shall now consider thebehaviour of f at points on J .
Definition 8 (ε−δ definition of continuity) Let f : J → R be given and a ∈ J . We say thatf is continuous at a if for any given ε > 0, there exists δ > 0 such that x ∈ J and |x−a| < δ⇒ |f(x)− f(a)| < ε.
Definition 9 A function f(x) is said to be continuous at a point c ∈ J , if limx→c
f(x) exists
and the limit equals to the value of the function at x = c (i .e., limx→c
f(x) = f(c)).
7
A function f is said to be continuous in an interval J , if it continuous at every point of theinterval.
A function is said to be discontinuous at a point x = c of its domain, if it is not continuousat x = c. The point x = c is called a point of discontinuity of the function.
Types of discontinuities:
(i) A function f is said to be have a removable discontinuity at x = c, if limx→c
f(x) exists
but is not equal to the value f(c) (which may or may not exist) of the function. Such adiscontinuity can be removed by assigning a suitable value to the function at x = c.(ii) A function f is said to have a discontinuity of the first kind at x = c, if lim
x→c−f(x) and
limx→c+
f(x) both exist but are not equal.
(iii) A function f is said to have a discontinuity of the first kind from the left at x = c, iflimx→c−
f(x) exists but is not equal to f(c).
(iv) A function f is said to have a discontinuity of the first kind from the right at x = c, iflimx→c+
f(x) exists but is not equal to f(c).
(v) A function f is said to have a discontinuity of the second kind at x = c, if neitherlimx→c−
f(x) nor limx→c+
f(x) exists.
(vi) A function f is said to have a discontinuity of the second kind from the left at x = c, iflimx→c−
f(x) does not exist.
(vii) A function f is said to have a discontinuity of the second kind from the right at x = c,if lim
x→c+f(x) does not exist.
Theorem 3 (without proof) Let f, g : J → R be continuous at a point a ∈ J . Let α ∈ R.Then the functions αf, |f |, f + g, f − g, fg are also continuous at x = a and if g(a) = 0, thenf/g is also continuous at x = a.
Definition 10 Let f, g : J → R be a real-valued function and a ∈ J . Assume that f iscontinuous at a and g is continuous at f(a). Then the composition function (gof) is alsocontinuous at a.
Definition 11 Let J ⊂ R. Let f : J → R be a real-valued function and a ∈ J . We say thatf is continuous at a if for every sequence {xn} in J with xn → a, we have f(xn) → f(a).
Example 16 Discuss the continuity of the following functions:(i) f(x) = 1/x, (ii) f(x) = sin(1/x).
Sol. (i) The function f(x) = 1/x has the natural domain R\{0}.Here
limx→0+
f(x) = ∞ and limx→0+
f(x) = −∞.
Thus f(x) has a discontinuity of the second kind at x = 0.(ii) The function f(x) = sin(1/x) has the natural domain R\{0}.It has been observed from Q. 1 in Assignment- 2, that it does not have a limit as x → 0.In fact lim
x→0+f(x) and lim
x→0−f(x) do not exist as f(x) oscillates between 1 and −1, as x → 0.
Thus f(x) has a discontinuity of the second kind at x = 0.
8
Example 17 Let
f(x) =
{x sin(1/x), if x = 0,0, if x = 0.
Show that f is continuous at x = 0.
Sol. Since |x sin(1/x)| ≤ |x|, it follows from the Lemma 1 (Sandwich Theorem) that
limx→0
x sin(1/x) = 0.
Thus limx→0
f(x) = f(0).
Hence the function f is continuous at x = 0.
Example 18 Discuss the continuity of
f(x) =
1 + x, −∞ < x < 0,1 + [x] + sinx, 0 ≤ x < π/2,3, x ≥ π/2.
Sol. Since f(0) = 1 = limx→0−
f(x) = limx→0−
(1 + x) = 1 = limx→0+
f(x) = limx→0+
(1 + [x] + sin x) = 1.
It follows that f is continuous at x = 0.Since [x] is right continuous but not left continuous at x = 1, so also is f .Again f(π/2) = 3 = lim
x→π/2+f(x),
limx→π/2−
f(x) = limx→π/2−
(1 + [x] + sin x) = 1 + limx→π/2−
[x] + limx→π/2−
sin x = 1 + 1 + 1 = 3.
Hence f is also continuous at π/2. Thus f is continuous at every point of R except at x = 1.
Example 19 Discuss the kind of discontinuity (if any) of the function defined as follows:
f(x) =
{x− |x|
x, x = 0,
2, x = 0.
Sol. Now limx→0−
f(x) = limx→0−
x+ x
x= 2,
limx→0+
f(x) = limx→0+
x− x
x= 0,
and f(0) = 2.Thus the function has discontinuity of the first kind from the right at x = 0.
Example 20 Show that the function defined by f(x) =
xe1/x
1 + e1/x, x = 0,
0, x = 0,is continuous
at x = 0.
Sol. Now limx→0−
f(x) = limx→0−
xe1/x
1 + e1/x= 0,
limx→0+
f(x) = limx→0+
x
e−1/x + 1= 0,
and f(0) = 0.∴ lim
x→0f(x) = 0 = f(0).
Thus the function is continuous at x = 0.
9
Example 21 Discuss the continuity of
f(x) =
{sin 2x
x, x = 0,
1, x = 0,
at x = 0.
Sol. Now limx→0
f(x) = limx→0
sin 2x
2x.2 = 2, so that lim
x→0f(x) = f(0).
Hence the limit exists, but is not equal to the value of the function at the origin.Thus the function has a removable discontinuity at x = 0.
Example 22 Let f : R → R be given by f(x) =
{1, x ∈ Q,0, x ∈ R−Q.
Then f is not contin-
uous at any point of R. This is known as Dirichlet’s function.
Sol. Let a ∈ Q so that f(a) = 1.Since in any interval there lie an infinite number of rational and irrational numbers, thereforefor each positive integer n, we can choose an irrational number an such that |an − a| < 1
n.
Thus the sequence {an} converges to a.But f(an) = 0 for all n and f(a) = 1, so that the sequence f(an) does not converge to
f(a) (i.e., limx→∞
f(an) = f(a)). Thus we conclude that f is not continuous at all a ∈ Q.
Next, let b ∈ R−Q. For each positive integer n we can choose a rational number bn suchthat |bn − b| < 1
n. Thus the sequence {bn} converges to b.
But f(bn) = 1 for all n and f(b) = 0, so that the sequence f(bn) does not converge tof(b) (i.e., lim
x→∞f(bn) = f(b)). Thus we conclude that f is not continuous at all b ∈ R−Q.
∴ f is continuous nowhere on R.
Assignment 3
1. Discuss the continuity and classify the discontinuities, if any, of the following functions;
(i) f(x) =
{x sin(1/x), x = 0,0, x = 0,
(ii) f(x) =
sin(x− c)
x− c, x = c,
0, x = c,
(iii) f(x) =
{x2, x ≥ 0,x, x < 0,
(iv) f(x) =1
x− acosec
1
x− a,
(v) f(x) =
{(1 + x)1/x, x = 0,1, x = 0,
(vi) f(x) =
e1/x
1− e1/x, x = 0,
1, x = 0,
(vii) f(x) =
{x, x ∈ Q,−x, x ∈ R−Q.
(viii) f(x) =
e1/x − e−1/x
e1/x + e−1/x, x = 0,
1, x = 0,
(ix) f(x) =
e1/x − 1
e1/x + 1, x = 0,
0, x = 0,(x) f(x) =
0, x = 0,(1/2)− x, 0 < x < 1/2,1/2, x = 1/2,3/2− x, 1/2 < x < 1,1, x = 1,
(xi) f(x) =
x3 − 8
x2 − 4, x = 2,
3, x = 2,(xii) f(x) =
{2x, x ≤ 1,x2, x > 1.
10
3 Differentiability of functions
Definition 12 Let J be an interval and c ∈ J . Let f : J → R. Then f is said to bedifferentiable at c, if there exists a real number α such that
limx→c
f(x)− f(c)
x− c= α. (1)
It is sometimes useful to use the variable h for the increment x − c and reformulate (1)as follows:
limh→0
f(c+ h)− f(c)
h= α. (2)
The limit value α is called the derivative of the function f at x = c and is denoted by f ′(c).
It should be noted that before examining the differentiability of f at c, it is necessary toensure that f is defined in a neighbourhood of c.
Let J be an open interval and let f : J → R be differentiable at every point of J , thenwe can define a new function f ′ : J → R defined by
f ′(x) = limy→x
f(y)− f(x)
y − x.
Thus we get an operator D (say), which takes f to f ′ whenever f is differentiable. ThusDf = f ′.
3.1 Left and right derivative
Definition 13 Let f : J → R, c ∈ J . The function f is said have a left derivative at thepoint x = c if there exists m ∈ R such that
limx→c−0
f(x)− f(c)
x− c= m,
and m is said to be left derivative of f at c and we denote it as f ′(c − 0), f ′(c−), Lf ′(c) orD−f(c).
Similarly, the function f is said have a right derivative at the point x = c if there existsm0 ∈ R such that
limx→c+0
f(x)− f(c)
x− c= m0,
and m0 is said to be right derivative of f at c and we denote it as f ′(c+0), f ′(c+), Rf ′(c) orD+f(c).
NOTE: If a function is differentiable at x = c, then(i) D−f(c) should exist.(ii) D+f(c) should exist.(iii) D−f(c) = D+f(c) = Df(c).Thus a function f is not differentiable if any one of the above requirements is not met.
In view of the definition of limit, the differentiability condition in (1) can be defined usingε− δ as follows:
11
Definition 14 We say that f is differentiable at c if there exists α ∈ R such that for anygiven ε > 0, there exists a δ > 0 such that
x ∈ J and 0 < |x− c| < δ ⇒ |f(x)− f(c)− α(x− c)| < ε|x− c|. (3)
We say that f is differentiable on J if it is differentiable at each c ∈ J .
Example 23 Let f : J → R be a constant function, say, C. Then f is differentiable atc ∈ J with f ′(c) = 0 by using ε− δ definition.
Sol. Consider the expression
f(x)− f(c)
x− c=
C − C
x− c= 0.
This suggests that α = f ′(c) = 0. Now we will prove that f ′(c) = 0 by using ε− δ definition.Let ε > 0 be given. Now let us try to estimate the error term:
|f(x)− f(c)− α(x− c)| = |C − C − α(x− c)| = |α||(x− c)| = 0.
This suggests that we can choose any δ > 0 for any ε > 0.Let ε > 0 be given. Let δ > 0 be arbitrary. We can estimate the error term:
|f(x)− f(c)− α(x− c)| = 0 < ε|x− c|.
Thus if f is a constant function then it is differentiable in R with f ′(c) = 0 for allc ∈ J ⊂ R.
Example 24 Let f : J → R be given by f(x) = ax + b. Then f is differentiable on R withf ′(c) = a, c ∈ R by using ε− δ definition.
Sol. Let c be an arbitrary real number.Consider the expression
f(x)− f(c)
x− c=
(ax+ b)− (ac+ b)
x− c=
a(x− c)
x− c= a.
This suggests that α = f ′(c) = a. Now we will prove that f ′(c) = a by using ε− δ definition.Let ε > 0 be given. Now let us try to estimate the error term:
|f(x)− f(c)− α(x− c)| = |a(x− c)− a(x− c)| = 0.
This suggests that we can choose any δ > 0 for any ε > 0.Let ε > 0 be given. Let δ > 0 be arbitrary. We can estimate the error term:
|f(x)− f(c)− α(x− c)| = 0 < ε|x− c|.
Since c is arbitrary real number, f is differentiable on R and f ′(c) = a.
Example 25 Find the derivative of the function f(x), where (i) f(x) = eαx, α ∈ R, (ii)f(x) = loga x.
12
Sol. (i) Given f(x) = eαx. Then f(x+ h)− f(x) = eα(x+h) − eαx = eαx(eαh − 1).Hence
limh→0
f(x+ h)− f(x)
h= eαx lim
h→0
eαh − 1
h
= eαx limh→0
1
h
[αh+
(αh)2
2!+ · · ·
]= αeαx.
(ii) Given f(x) = loga x. Thenf(x+ h)− f(x)
h=
loga(x+ h)− loga x
h= loga
(1 +
h
x
)1/h
.
Hence
limh→0
f(x+ h)− f(x)
h= lim
h→0loga
(1 +
h
x
)1/h
= loga limh→0
(1 +
h
x
)1/h
= loga e1/x
= (1/x) loga e.
Example 26 Discuss the differentiability of the following functions:
(i) f(x) =
{x, 0 ≤ x < 1,1, x ≥ 1.
at x = 1 and (ii) f(x) = x2 on the interval [0, 1].
Sol. (i) Given that f(x) =
{x, 0 ≤ x < 1,1, x ≥ 1.
Now
D−f(1) = limx→1−0
f(x)− f(1)
x− 1= lim
x→1−
x− 1
x− 1= 1,
D+f(1) = limx→1+0
f(x)− f(1)
x− 1= lim
x→1+
1− 1
x− 1= 0.
∴ D−f(1) = D+f(1).
Thus f ′(1) = limx→1
f(x)− f(1)
x− 1does not exist. Hence f is not differentiable at x = 1.
(ii) Given that f(x) = x2. Let x0 be any point on (0, 1), then
f ′(x0) = limx→x0
f(x)− f(x0)
x− x0
= limx→x0
x2 − x20
x− x0
= limx→x0
(x+ x0) = 2x0.
At the end points, we have
f ′(0) = limx→0+0
f(x)− f(0)
x− 0= lim
x→0+
x2
x= lim
x→0+x = 0,
and
f ′(1) = limx→1−0
f(x)− f(1)
x− 1= lim
x→1−
x2 − 1
x− 1= lim
x→1−x+ 1 = 2,
Thus the function f is derivable in the closed interval [0, 1].
Theorem 4 (without proof) A function which is differentiable at a point is continuous atthe point; but not conversely.
13
Theorem 5 (without proof) Let f, g : J → R be differentiable at c ∈ J . Then the followinghold:(i) f + g is differentiable at c with (f + g)′(c) = f ′(c) + g′(c),(ii) αf is differentiable at c with (αf)′(c) = αf ′(c),(iii) fg is differentiable at c with (fg)′(c) = f ′(c)g(c) + f(c)g′(c),
(iv) f/g is differentiable at c with (f/g)′(c) =f ′(c)g(c)− f(c)g′(c)
{g(c)}2, provided g(c) = 0,
(v) (gof) is differentiable at c with (gof)′(c) = g′(f(c))f ′(c).
Example 27 The function f(x) = |x| is continuous on R but not differentiable at x = 0.
Sol. Given that f(x) = |x|, ∀x ∈ R.Now
limx→0+
f(x) = limx→0−
f(x) = 0 = f(0).
Thus f is continuous at x = 0.We can easily check that f is continuous for all x > 0 and x < 0.Thus f is continuous for all x ∈ R.But
D+f(0) = limx→0+
|x| − 0
x− 0= lim
x→0+
x
x= lim
x→0+1 = 1.
and
D−f(0) = limx→0−
|x| − 0
x− 0= lim
x→0−
−x
x= lim
x→0−(−1) = −1.
Thus D+f(0) = D−f(0).Hence the function f is not differentiable at x = 0.
Example 28 The function f(x) =
{x sin(1/x), x = 0,0, x = 0,
is continuous but differentiable
at x = 0.
Sol. This function is continuous at x = 0 (see Example-17).But since
f(x)− f(0)
x− 0=
x sin(1/x)
x= sin(1/x),
it does not possess left or right limit at x = 0. Hence f is not differentiable at x = 0.
Example 29 The function f(x) =
{x2 sin(1/x), x = 0,0, x = 0,
is differentiable at x = 0 but f ′
is not continuous at x = 0 (i.e., limx→0
f ′(x) = f ′(0)).
Sol. We have
Df(0) = f ′(0) = limx→0
f(x)− f(0)
x− 0= lim
x→0
x2 sin(1/x)
x= lim
x→0x sin(1/x) = 0.
Thus f differentiable at x = 0 and f ′(0) = 0.If x = 0, then from the elementary calculus, we know that
f ′(x) = 2x sin(1/x)− cos(1/x). (4)
Clearly, limx→0
f ′(x) does not exist as cos(1/x) oscillates at x = 0 and therefore there is no
possibility of limx→0
f ′(x) being equal to f ′(0).
Thus f ′(x) is not continuous at x = 0.Note further from equation (4) that f ′ is not differentiable at x = 0.
14
Assignment 4 Test the differentiability of the following functions at the points indicated.
(i) f(x) =
{sinx, x ≤ π/2,1 + (x− π/2)2, x > π/2,
at the point x = π/2
(ii) f(x) =
{e1/(x
2−1), |x| < 1,0, |x| ≥ 1,
at the points x = −1 and x = 1.
(iii) f(x) =
{e1/(x
2−1), x < 1,0, x ≥ 1,
at the point x = 1.
(iv) f(x) = [x](x− 1) at x = 0 and x = 1.(v) f(x) = |x|+ |x− 1| at x = 0 and x = 1.
(vi) f(x) =
{e−1/x2
, x > 0,0, x ≤ 0,
at the point x = 0.
(vii) f(x) =
{2x− 3, 0 ≤ x ≤ 2,x2 − 3, 2 < x ≤ 4,
at the points x = 2 and x = 4.
(viii) f(x) =
x(e1/x − 1)
e1/x + 1, x = 0,
0, x = 0,at the point x = 0.
(ix) f(x) =
x(e1/x − e−1/x)
e1/x + e−1/x, x = 0,
0, x = 0,at the point x = 0.
(x) f(x) =
{x tan−1(1/x), x = 0,0, x = 0,
at the point x = 0.
Theorem 6 Darboux Theorem (without proof)Let f : [a, b] → R be differentiable. Assume that f ′(a) < λ < f ′(b). Then there exists c ∈(a, b) such that f ′(c) = λ. (Thus though f ′ need not be continuous, it enjoys the intermediatevalue property.)
Intermediate value theorem for derivatives- If a function f is derivable on a closedinterval [a, b] and f ′(a) = f ′(b) and λ a number lying between f ′(a) and f ′(b), then thereexists at least one point c ∈ (a, b) such that f ′(c) = λ.
Example 30 Define f(x) =
{x2 sin(1/x), x = 0,0, x = 0.
Then f is differentiable at all points including 0.
Here f ′(x) =
{2x sin(1/x)− cos(1/x), x = 0,0, x = 0.
It is easy to see that f ′ is not continuous (see Example-29).According to Darboux theorem, f ′ enjoys the intermediate value property, even though
it is not continuous.
4 Mean Value Theorems
On the basis of a certin amount of knowledge about the derivative of a function, mean valuetheorems enable us to get some information about the function itself. Sometimes, it is easierto tackle the derivative than the function. In this section we shall establish the so-called firstderivative test.
15
Theorem 7 Rolle’s Theorem (without proof)Let f : [a, b] → R be such that (i) f is continuous on [a, b], (ii) f is differentiable on (a, b),and (iii) f(a) = f(b). Then there exists at least one real point c ∈ (a, b) such that f ′(c) = 0.
The geometric interpretation is that there exists at least one point c ∈ (a, b) such that slopeof the tangent to the graph of f at c equals to zero. That is, the tangent at (c, f(c)) is parallelto the x-axis.There is also an algebraic interpretation of Rolle’s Theorem. If f(a) = f(b) = 0, then a andb are the zeros of f(x) or the roots of the equation f(x) = 0. Thus, Rolle’s Theorem saysthat if a and b are two roots of the equation f(x) = 0, then there exists at least one rootc ∈ (a, b) of the equation f ′(x) = 0.
Example 31 Show that the equation 10x4 − 6x+ 1 = 0 has a root between 0 and 1.
First determine the polynomial function f whose derivative is the polynomial, whose rootsare being sought. So we take f(x) = 2x5 − 3x2 + x.
It is easily seen that f is continuous on [0, 1] and differentiable on (0, 1). Also f(0) = 0 =f(1).
Hence using the Rolle’s theorem, there exists c ∈ (0, 1) such that f ′(c) = 10c4−6c+1 = 0.
Example 32 Prove that the function f(x) = x3 + x + k = 0, k in any real constant, hasexactly one real root.
If it has two distinct roots, say c1 and c2, then f(c1) = 0 and f(c2) = 0.Again, any polynomial function is continuous and differentiable on R. Hence by Rolle’s
theorem there exists c ∈ (c1, c2) such that f ′(c) = 3c2 + 1 = 0, which contradicts the factthat c ∈ R.
Hence f has exactly one real root on R.
Theorem 8 Mean Value Theorem (MVT) or Langrange’s Mean Value Theorem (withoutproof)Let f : [a, b] → R be such that (i) f is continuous on [a, b] and (ii) f is differentiable on(a, b). Then there exists at least one real number c ∈ (a, b) such that
f(b)− f(a) = (b− a)f ′(c).
There is a geometric interpretation of Mean Value Theorem. Under the given conditions,there exists c ∈ (a, b) such that the slope of tangent to the graph of f at c equals that of thechord joining the two points (a, f(a)) and (b, f(b)).
Theorem 9 Cauchy’s form of Mean Value Theorem (without proof)Let f, g : [a, b] → R be such that (i) f, g are continuous on [a, b], (ii) f, g are differentiableon (a, b), and (iii) g′(x) = 0, for any x ∈ (a, b). Then there exists at least one real numberc ∈ (a, b) such that
f(b)− f(a)
g(b)− g(a)=
f ′(c)
g′(c).
Geometrically, Cauchy’s form of MVT (Mean Value Theorem) means the following:We look at the map t 7→ (g(t), f(t)) from J to R2 as a parameterized curve in the plane. Forexample, t 7→ (cos t, sin t), t ∈ [0, 2π] is parameterization of a circle.
Then the slope of the chord joining the points (g(a), f(a)) and (g(b), f(b)) is f(b)−f(a)g(b)−g(a)
.
The tangent vector to the parameterized curve at a point (g(t0), f(t0)) is (g′(t0), f
′(t0)) and
16
hence the tangent line at t0 has the slope f ′(t0)/g′(t0). Thus Cauchy’s mean value theorem
says that there exists a point c ∈ (a, b) such that slope f ′(c)/g′(c) of the tangent to the curveat c is equal to the slope of the chord joining the end points of the curve.
As observed in the remark on the geometric interpretation, if g(x) = x in the Cauchy’smean value theorem, it reduces to Lagrange’s mean value theorem.
Theorem 10 Applicatons of Mean Value Theorem (MVT)Let f : [a, b] → R be differentiable on (a, b).(i) If f ′(x) > 0 for all x ∈ (a, b), then f is strictly increasing on (a, b),(ii) If f ′(x) = 0 for all x ∈ (a, b), then f is a constant on (a, b),(iii) If f ′(x) < 0 for all x ∈ (a, b), then f is strictly decreasing on (a, b).
The mean value theorem is quite useful in proving certain inequalities. Here are somesamples.
Example 33 Show that ex > 1 + x for all x ∈ R\{0}.
Suppose x > 0. Consider the function f(x) = ex on the interval [0, x]. Since ex is differentiableon R, we can apply mean value theorem to f on the interval [0, x]. Hence there exists c ∈ (0, x)such that
ex − e0 = f ′(c)(x− 0) = ecx.
Note that f ′(x) = ex > 1 for all x > 0. So the displayed equation gives ex − 1 = ecx > x.Similarly if x < 0, then consider the interval [x, 0] and we can prove that ex > 1 + x.
Example 34 Prove thaty − x
y< log
y
x<
y − x
x, 0 < x < y.
Let 0 < x < y and f(x) = log x on [x, y]. We know that log x is differentiable function onx > 0. Hence using the MVT, there exists c ∈ (x, y) such that
log y − log x = f ′(c)(y − x) ⇒ logy
x=
1
c(y − x).
Since 0 < x < c < y, we have1
y<
1
c<
1
x. Hence we get
y − x
y<
1
c(y − x) = log
y
x<
y − x
x.
Example 35 Prove thatsinα− sin β
cos β − cosα= cot θ, for 0 < α < θ < β < π/2.
Let f(x) = sinx and g(x) = cosx for x ∈ [α, β].∴ f ′(x) = cos x and g′(x) = − sinx.Here the functions f and g are both continuous and differentiable. Therefore, by Cauchy’s
mean value theorem on [α, β],
sin β − sinα
cos β − cosα=
cos θ
− sin θ, α < θ < β,
or,sinα− sin β
cos β − cosα= cot θ, α < θ < β.
17
Example 36 Prove thatx
1 + x< log(1 + x) < x for all x > 0.
Let f(x) = x − log(1 + x). Hence f ′(x) = 1 − 1
1 + x=
x
1 + x> 0. So by Theorem 10, f is
strictly increasing. Since f(0) = 0, f(x) > 0 for x > 0. Thus x > log(1 + x).
Similarly we consider the function g(x) = log(1 + x)− x
1 + xand show that g(x) > 0 for
x > 0.
Example 37 Show thattanx
x>
x
sin xfor 0 < x < π/2.
We have to show thattanx
x− x
sin x> 0, or
sinx tanx− x2
x sin x> 0 for 0 < x < π/2. Since
x sin x > 0 for 0 < x < π/2, therefore it will sufficient to show that sin x tanx− x2 > 0.Let f(x) = sin x tanx − x2, then for 0 < x < π/2, f ′(x) = sinx + sin x sec2 x − 2x. We
cannot decide about the sign of f ′(x) mainly because of the presence of the (−2x) term. Nowthe function f ′(x) is continuous and derivable on (0, π/2).
∴ f ′′(x) = cos x+ cos x sec2 x+ 2 sinx sec2 x tanx− 2
= (√secx−
√cosx)2 + 2 tan2 x sec x > 0, for 0 < x < π/2.
Since the derivative f ′′(x) of f ′(x) is positive, the function f ′(x) is an increasing function.Further since f ′(0) = 0, therefore the function f ′(x) > 0 for 0 < x < π/2.
Again, since f ′(x) > 0, f(x) is an increasing function and because f(0) = 0, the functionf(x) > 0, for 0 < x < π/2.
Thus it follows thattanx
x>
x
sinxfor 0 < x < π/2.
Assignment 5 Solve the following problems by using Rolle’s theorem/ MVT;1. Prove that between any two real roots of ex sin x = 1, there is at least one real root ofex cosx+ 1 = 0.2. Show that the equation cos x = x3 + x2 + 4x has exactly one root in [0, π/2].3. Prove that the equation x3 − 3x2 + b = 0, b ∈ R has at most one r 4. Let f : [2, 5] → R becontinuous and be differentiable on (2, 5). Assume that f ′(x) = (f(x))2+π for all x ∈ (2, 5).True or false (with proper reason): f(5)− f(2) = 3.5. Apply Lagrange’s mean value theorem for the function f(x) = log(1 + x) to show that
0 <1
log(1 + x)− 1
x< 1, for all x > 0.
6. Establish the following inequalities:
(i) x− x2
2+
x3
3(1 + x)< log(1 + x) < x− x2
2+
x3
3, x > 0,
(ii)x2
2(1 + x)< x− log(1 + x) <
x2
2, x > 0,
(iii)x2
2< x− log(1 + x) <
x2
2(1 + x), −1 < x < 0,
(iv) (1 + x) < ex < 1 + xex, for all x,
(v) (1− x) < e−x < 1− x+x2
2, for all x > 0.
18
Theorem 11 Taylor’s Theorem (without proof)Assume that f : [a, b] → R be such that f (n) is continuous on [a, b] and f (n+1)(x) exists on(a, b). Fix x0 ∈ [a, b]. Then for each x ∈ [a, b] with x = x0, there exists c between x and x0
such that
f(x) = f(x0) +n∑
k=1
(x− x0)k
k!f (k)(x0) +Rn, (5)
where Rn = (x−x0)n+1
(n+1)!f (n+1)(c).
The right-hand side of (5) is called the n-th order (or n-th degree) Taylor expansion of
the function f at x0. The expression f(x) = f(x0) +n∑
k=1
(x− x0)k
k!f (k)(x0), is called the
n-th degree Taylor polynomial of f at x0. The term Rn is called the remainder term in theTaylor’s expansion after (n + 1) terms. The remainder term is the “error term” if we wishto approximate f near x0 by the n-th order Taylor polynomial. If we assume that f (n+1) isbounded, say, by M on (a, b), then Rn goes to zero much faster than (x− x0)
n → 0.Putting x0 = 0 in expression in (5) is called Maclaurin’s Theorem with form of remainder.
Example 38 Show that tha function f(x) = sin x, x ∈ [0, π/4] is approximated by a polyno-mial sinx = x− (x3/6), with an error less than 1/400.
The function f satisfies the condition of Taylor’s theorem. Hence it can be expressed atx0 = 0 by
f(x) = f(0) + xf ′(0) +x2
2!f ′′(0) +
x3
3!f ′′′(0) +
x4
4!f iv(0) +R5,
where for some c ∈ (0, x), and R5 =x5
5!f (v)(c). Thus
f(x) = x− x3
6+
x5
120cos c. (6)
Now
|R5| =x5
120| cos c| ≤ x5
120≤ 1
120(π/4)5 <
1
400. (7)
Hence it follow from (6) and (7) that
|f(x)− p(x)| < 1
400, where p(x) = x− x3
6.
Assignment 6 1. If 0 < x ≤ 2, then prove that
log x = (x− 1)− (x− 1)2
2+
(x− 1)3
3− (x− 1)4
4+ · · · .
2. Assuming the validity of expansion, show that
(i) ex cos x = 1 + x− 2x3
3!− 22x4
4!− 22x5
5!+ · · · .
(ii) log sec x =1
2x2 +
1
12x4 + · · · .
(iii) tan−1 x = tan−1 π
4+
x− π/4
1 + π2/16− π(x− π/4)2
4(1 + π2/16)2+ · · · .
(iv) sin(π
4+ θ) =
1√2
(1 + θ − θ2
2!− θ3
3!+ · · ·
).
19
References
[1] A. Kumar and S. Kumaresan, A basic course in Real Analysis, CRC Press, 2014.
[2] G. Das and S. Pattanayak, Fundamentals of Mathematical Analysis, Tata McGraw HillPublication, 2010.
[3] S.C. Malik and S. Arora, Mathematical Analysis, New Age International Publication,2008.
20
Lecture Notes
Mathematics-I, B.Tech, First Semester
VSSUT, Burla
Odisha 768018
———————————————————————————————————
——————–
UNIT-II
1 Linear Algebra
1.1 Matrix
A matrix is a rectangular array of numbers (or functions) enclosed in brackets.
Example 1.1.
1 2 3 4
A a1 a2 a3 a4
B b1 b2 b3 b4
C c1 c2 c3 c4
D d1 d2 d3 d4
or
a1 a2 a3 a4
b1 b2 b3 b4
c1 c2 c3 c4
d1 d2 d3 d4
or
1
a1 a2 a3 a4
b1 b2 b3 b4
c1 c2 c3 c4
d1 d2 d3 d4
ora1 a2 a3 a4
b1 b2 b3 b4
or{f1(x) f2(x) f3(x) f4(x)
}
Vector:
A vector is a matrix that has only one row- we call row matrix- or only one
column we called column matrix. Entries of matrix are called components.
Example 1.2.{a1 a2 a3 a4
}or
a1
b1
c1
d1
Transposition:
Given a matrix
2
A=[ajk]=
a11 a12 a13 a14 . . a1n
a21 a22 a23 a24 . . a2n
a31 a32 a33 a34 . . a3n
. . . . . . .
am1 am2 am3 am4 . . amn
. The transpose of A is written as AT
is an n×m matrix defined as follows
AT=[akj]=
a11 a21 a31 . . . am1
a12 a22 a32 . . . a2n
a13 a23 a33 . . . a3n
. . . . . . .
a1n a2m a3m . . . amn
Symmetric and Skew symmetric, Equal matrix:-
A matrix A is called symmetric, if A = AT and is called skew symmetric matrix
if AT = −A , two matrices are equal if they have same size and corrsponding
entries are equal.
Matrix Addition:- Addition of matrices A=[ajk] and B=[ail] is defined only
when both are of same size, their sum A+B is obtained by adding the corrspond-
ing entries.
Example 1.3.
3 4 −1
4 −5 7
0 1 −1
+
0 −2 −1
3 −5 2
0 5 −2
=
3 2 −2
7 −10 9
0 6 −3
Scalar Multiplication: The product of any m× n matrix A = [ajk] and any
scalar c written as cA, is the m× n matrix cA = [cajk]
Matrix multiplication: Let A = [ajk] is an m × n matrix and B = [ajk] is an
r×p matrix, then product C = AB is defined if and only if r = n, with C = [cjk]
3
where
cjk =n∑i=1
ajibik = aj1b1k + aj2b2k + · · · ajnbnk j = 1.2...m and k = 1.2...p
Example 1.4.
AB =
4 3
7 2
9 0
2 5
1 6
=
11 38
16 47
18 45
Some Properties of matrix multiplication
1. (kA)B = k(AB) = A(kB), where K is any scalar.
2. A(BC) = (AB)C.
3. (A+B)C = AC +BC.
4. AC = AD ; C = D, even when A 6= 0.
5. Matrix multiplication is not commutative in general.
Special Matrices: We now list here some of the important matrices: Trangular
Matrices
Upper triangular matrices are square matrices that can have nonzero entries
only on and above the main diagonal where as any entry below the diagonal must
be zero. Similarly lower triangular matrices can have nonzero entries only on and
bellow the main diagonal.
Example 1.5.
Upper and Lower triangular matrices
1 5
0 −5
,
7 3 2
0 6 1
0 0 3
4
1 0 0
7 8 0
8 −2 1
The first two matrices are upper triangular whereas the last one is the lower
triangular matrix.
Diagonal matrices The square matrices whose main diagonal entries are
nonzero are called diagonal matrices.
Example 1.6.
diagonal matrices:
1 0 0
0 8 0
0 0 7
,
14 0 0
0 86 0
0 0 57
Transpose of a product: The transpose of a product equals the product of
the transposed factors, taken in reverse order,
(AB)T = BTAT .
Inner product of vectors: If a is a row vector and b is column vector both
with n components then the inner product or dot product of aand b is defined by
a.b = [a1, a2, · · · , an]
b1
b2
...
bn
=
n∑i=1
aibi = a1b1 + a2b2 + · · ·+ anbn.
5
Homework
a =
1
4
3
, B =
2 −3
0 2
0 1
, C =
4 6 2
6 0 3
2 3 −1
, d =
(4 3 0
)
Calculate the following products.
1. Ba, aTB, aB.
2. C2, CTC, CCT .
3. aTd, BTB, da, ad.
4. Ca, C2a, C3a.
Linear system of equations, Gauss Elimination: A linear system of m-
equations in n- unknowns x1, x2...xn, is a set of equations of the form
a11x1 + a12x2 + a13x3 + ...+ a1nxn = b1
a21x1 + a22x2 + a23x3 + ...+ a2nxn = b2
.......................................................
am1x1 + am2x2 + am3x3 + ...+ amnxn = bm (1)
A solution of (1) is a set of numbers x1, x2...xn, that satisfy all the m equations.
6
Matrix form of linear equation: Ax = b, where A is the coefficient matrix,
A =
a11 a12 a13 a14 . . a1n
a21 a22 a23 a24 . . a2n
a31 a32 a33 a34 . . a3n
. . . . . . .
am1 am2 am3 am4 . . amn
and x =
x1
x2
x3
.
.
.
xn
, b =
b1
b2
b3
.
.
.
bm
If all bi = 0 then (1.1) is called homogeneous system. If at least one bi 6= 0 then
the system is called non homogeneous.
The matrix
A =
a11 a12 a13 a14 . . a1n b1
a21 a22 a23 a24 . . a2n b2
a31 a32 a33 a34 . . a3n b3
. . . . . . .
am1 am2 am3 am4 . . amn bm
is called augmented matrix.
Theorem 1.7. Row equivalent linear systems have the same sets of solution.
Gauss Elimination Method:
Example 1.8.
7
Solve the linear system.
− x1 + x2 + 2x3 = 2
3x1 − x2 + x3 = 6
− x1 + 3x2 + 4x3 = 4
Solution 1.9.
−1 1 2 | 2
3 −1 1 | 6
−1 3 4 | 4
−1 1 2 | 2
3 2 7 | 12
0 2 2 | 2
R2 ← R2 + 3R1
R2 ← R3 − 3R1
−1 1 2 | 2
0 2 7 | 12
0 0 −5 | −10
R3 ← R2 − 3R2
The above row equivalent form gave a set of equation as follows
− x1 + x2 + 2x3 = 2
2x2 + 7x3 = 12
− 5x3 = 10
8
Solving the above system of equation we get
x1 = 2, x2 = −1, x3 = 1
Home work Solve the following systems by the Gauss-elimination method.
1.
6x+ 4y = 2
3x− 5y = −34.
2.
0.4x+ 1.2y = −2
1.7x− 3.2y = 8.1.
3.
13x+ 12y = −6
−4x+ 7y = −72
11x− 13y = 157.
4.
1.3x− 9.1y + 11.7z = 0
−0.9x+ 6.3y − 8.1z = 0
2 Rank of a matrix, Linear Independence:
Linear Independence and Dependence of Vectors Given any set of m
vectors (with the same number of components), a linear combination of these
vectors is an expression of the form
c1a1 + c2a2 + · · ·+ cmam
where c1, c2, · · · , cm are any scalars. Now consider the equation
c1a1 + c2a2 + · · ·+ cmam = 0 (2)
If all the scalars cj are zero, then our vectors are said to form a linearly inde-
pendent set or, more briefly, we call them linearly independent. Otherwise, if (2)
also holds with scalars not all zero, we call these vectors linearly dependent. This
means that we can express at least one of the vectors as a linear combination of
the other vectors.9
Example 2.1.
Linear Independence and Dependence The three vectors
a1 = [3, 0, 2, 2], a2 = [−6, 42, 24, 54] a3 = [21,−21, 0,−15]
are linearly dependent because
a3 = 6a1 −1
2a2.
Rank of a Matrix
Definition 2.2. The rank of a matrix A is the maximum number of linearly
independent row vectors of A. It is denoted by rank A.
Theorem 2.3. Row-equivalent matrices have the same rank.
Theorem 2.4. The rank of A equals the maximum number of linearly indepen-
dent column vectors of A. Hence A and AT have the same rank.
Theorem 2.5. Consider p vectors each having n components. If n < p then
these vectors are linearly dependent.
Example 2.6.
Find the rank of the matrix A =
3 0 2 2
−6 42 24 54
21 −21 0 −15
Solution 2.7. The given matrix A has rank 2 because first two row vectors are
linearly independent, whereas all three row vectors are linearly dependent.
Home work Find the rank of the matrix.
1.
1
4
3
,
2 −3
0 2
0 1
,
4 6 2
6 0 3
2 3 −1
,
(4 3 0
)
10
2.
1 0 0
0 8 0
0 0 7
,
14 0 0
0 86 0
0 0 57
3.
1 5
0 −5
,
7 3 2
0 6 1
0 0 3
4.
1 0 0
7 8 0
8 −2 1
3 Vector spaces and subspaces
Fields of scalars
Definition A field of scalars (or just a field) consists of a set F whose elements
are called scalars, together with two algebraic operations, addition + and mul-
tiplication ., for combining every pair of scalars x, y ∈ F to give new scalars
x+ y, x.y ∈ FDefinition A vector space over a field of scalars F consists of a set V whose
elements are called vectors together with two algebraic operations, + (addition of
vectors) and (multi- plication by scalars). Vectors will usually be denoted with
boldface symbols such as v ,The operations + and . are required to satisfy the
following rules, which are sometimes known as the vector space axioms.
Associativity: For u, v, w ∈ V ands, t ∈ F , (u+ v) + w = u+ (v + w);
(s.t).v = s.(t.v);
Zero and unity: There is a unique element 0 ∈ V such that for v ∈ V ,
v + 0 = v = 0 + v; and multiplication by 1 ∈ F satisfies 1.v = v:
Distributivity: For s, t ∈ F and u, v ∈ V , (s + t).v = s.v + t.v; s.(u + v) =
11
s.u+ s.u:
Commutativity: For u; v2V , u+ v = v + u:
Additive inverses: For v ∈ V there is a unique element −v ∈ V for which
v + (−v) = 0 = (−v) + v
Definition Sub Space: Let V be a vector space over a field of scalars F. Suppose
that the subset W of V is non-empty and is closed under addition and multipli-
cation by scalars i.e., for s ∈ F, u, v ∈ W u + v ∈ W and su ∈ W thus it forms
a vector space over F. Then W is called a (vector) subspace of V .
The maximum number of linearly independent vectors in V is called the dimen-
sion of V and is denoted by dim V. A linearly independent set in V consisting
of a maximum possible number of vectors in V is called a basis for V. In other
words, any largest possible set of independent vectors in V forms basis for V.
That means, if we add one or more vector to that set, the set will be linearly
dependent. Thus, the number of vectors of a basis for V equals dim V. The set
of all linear combinations of given vectors a1, a2, · · · , an with the same number
of components is called the span of these vectors. Obviously, a span is a vector
space. If in addition, the given vectors a1, a2, · · · , an are linearly independent,
then they form a basis for that vector space.
4 Inverse of a matrix, Gauss Jordan Elimina-
tion:
In this section we consider square matrices exclusively. The inverse of an n×nmatrix is denoted by A−1 and is an n× n matrix such that
AA−1 = A−1A = I
where I is the unit matrix. If A has an inverse, then A is called a nonsingular
matrix. If A has no inverse, then A is called a singular matrix. If A has an
inverse, the inverse is unique. Indeed, if both B and C are inverses of A, then
AB = I and CA = I so that we obtain the uniqueness from
B = IB = (CA)B = C(AB) = CI = C.
We prove next that A has an inverse (is nonsingular) if and only if it has maximum
possible rank n. The proof will also show Ax = b that implies x = A−1b provided
12
A−1 exists, and will thus give a motivation for the inverse as well as a relation to
linear systems.
Theorem 4.1. Existence of the Inverse
The inverse A−1 of an n × n matrix A exists if and only if rank A = n, thus
if and only if |A| 6= 0. Hence A is nonsingular if rank A = n and is singular if
rank A < n.
Determination of the Inverse by the GaussJordan Method To determine the in-
verse of a nonsingular matrix A, we can use a variant of the Gauss elimination
called the GaussJordan elimination. The idea of the method is as follows. Using
the matrix A, we form n linear systems
Ax1 = e1, Ax2 = e2, · · ·Axn = en.
where the vectors e1, e2, · · · en are the columns of the unit matrix I. These
are n vector equations in the unknown vectors x1, · · · , xn. We combine them
into a single matrix equation AX = I with the unknown matrix X having the
columns x1, x2, · · · xn. Correspondingly, we combine the n augmented matrices
[A, e1], · · · [A, en] into one wide n× 2n augmented matrix A = [AI] . Now multi-
plication of AX = I by A−1 from the left gives X = A−1I = A−1. Hence, to solve
AX = I for X, we can apply the Gauss elimination to A = [AI] . This gives a
matrix of the form [UH] with upper triangular U because the Gauss elimination
triangularizes systems. The GaussJordan method reduces U by further elemen-
tary row operations to diagonal form, in fact to the unit matrix I. This is done
by eliminating the entries of U above the main diagonal and making the diagonal
entries all 1 by multiplication.
Exercises Calculate the inverse by the Gauss-Jordan Elimination
1.
2 0 −1
5 1 0
0 1 3
13
2.
4 −1 −5
15 1 −5
5 4 9
3.
1 8 −7
0 1 3
0 0 1
14
UNIT- III
5 First order differential equation:
Definition 5.1. A differential equation is an equation involving derivatives of one
or more dependent variable with respect to one or more independent variable.
For example,
y′′ + 4y = 0
y′ + xy = x2.
y′′′ + x4y′′ + y′4 + y = cosx.
Definition 5.2. An ordinary differential equation is one which involves only
one independent variable, so that all the derivative occurring in the differential
equation are ordinary derivatives.
Solution of a differential equation: When we say x = 1 is a solution of
the algebraic equation x2 = 1 we mean that when x = 1 is substituted in the
equation the equality will hold. Similarly we say that y = x2 is a solution of the
differential equation dydx
= 2x since if we put y = x2 in the above equation the
equality holds. Thus we give the formal definition of the solution of a general
ordinary differential equation of nth order
F (x, y, y′, y′′, · · · y(n)) = 0 (3)
as follows
Definition 5.3. Let y = f(x) be a real valued function on a interval I, then
f is called an explicit solution of the differential equation (3) if the substitution
y = f(x) reduces to an identity in x on I, i.e., if
F (x, f(x), f ′(x), · · · , f (n)(x)) = 0
for every x in I.
Definition 5.4. A relation g(x, y) = 0 is called an implicit solution of the dif-
ferential equation (3) on I if this relation defines at-least one real function of
x, y = φ(x) on I such that φ is an explicit solution of (3) on I.
15
Example 5.5.
Verification of solution:
Verify that y = x2 is a solution of xy′ = 2y for all x.
Indeed by substituting y = x2 and y′ = 2x into the equation we obtained xy′ =
x(2x) = 2x2 = 2y, an identity in x.
Example 5.6.
The relation
x2 + y2 = 1 (4)
is implicit solution of the differential equation
x+ ydy
dx= 0 (5)
on the interval I : −1 < x < 1. To verify this, first we see that the relation (4)
defines two real functions.
φ1(x) = +√
1− x2
and , φ1(x) = −√
1− x2, for x ∈ I : −1 < x < 1.
Next we see that the real function φ1 is an explicit solution of (5), for sub-
stituting y = φ1(x) =√
1− x2 and y′ = φ′1(x) = −x√1−x2 In (5) we obtain
x +√
1− x2 −x√1−x2 = 0 which is an identity for all x ∈ I. Thus the relation
(4) is an implicit solution of (5).
Generally it is difficult to solve first order ordinary differential equations dydx
=
f(x, y) in the sense that no formulae exist for obtaining its solution in all cases.
However their are certain standard types of first order differential equations of
first degree for which routine methods of solution are available. In this unit we
shall discuss a few of these types.
6 Separable differential equation:
Differential equation of the form
g(y)dy = f(x)dx
are called equations with separated variables, the solutions of which are obtained
by direct integration. Thus its solution is given by∫g(y)dy =
∫f(x)dx+ c.
16
Example 6.1.
Solve the differential equation
x(1 + y2)dx− y(1 + x2)dy = 0.
Solution 6.2. Separating the variables by dividing the product of (1+y2)(1+x2)
we getydy
1 + y2=
xdx
1 + x2.
By integrating on both sides we obtain the general solution 1 + y2 = c(1 + x2.
Example 6.3.
Solve the differential equation
9yy′ + 4x = 0.
Solution 6.4. By separating the variables we have 9ydy = −4xdx. On integra-
tion, this yieldsx2
9+y2
4= c.
The solution represents a family of ellipses.
Exercise: Solve the following differential equation
1. yy′ + 25x = 0.
2. y′ + 3x2y2 = 0.
3. y′ = x2+y2
xy.
4. xydx+ (x+ 1)dy = 0.
5. sec2 x tan ydx+ sec2 y tanxdy = 0.
6. (x+ y)dx+ dy = 0.
7 Exact differential equations and Integrating
Factor:
Definition 7.1. The differential equation
M(x, y)dx+N(x, y)dy = 0 (6)17
is called exact if there exist a function u(x, y) such that du(x, y) = M(x, y)dx +
N(x, y)dy
Theorem 7.2. The differential equation
M(x, y)dx+N(x, y)dy = 0
is exact if and only if∂M
∂y=∂N
∂x.
A working rule to find out the solution of the exact differential equation
M(x, y)dx+N(x, y)dy = 0
is as follows:∫y as constant
M(x, y)dx+
∫only those terms which do not contain x
N(x, y)dy = c
Example 7.3.
Solve
(x3 + 3xy2)dx+ (3x2y + y3)dy = 0.
Solution 7.4. Here M(x, y) = (x3 + 3xy2) and N(x, y) = (3x2y + y3). Thus
∂M
∂y= 6xy,
∂N
∂x= 6xy.
Since∂M
∂y=∂N
∂x
the given equation is exact. Hence the solution of the differential equation is∫(x3 + 3xy2)dx+
∫(3x2y + y3)dy = c.
In the first integration we treat y as constant and in the second we will integrate
only those terms which do not contain x. That implies
1
4(x4 + 6x2y2 + y4) = c
is the desired solution.
Example 7.5.
Test the equation
eydx+ (xey + 2y)dy = 0
for exactness, and solve it if it is exact.
18
Solution 7.6. Here M(x, y) = ey and N(x, y) = xey + 2y. Therefore
∂M
∂y= ey, and
∂N
∂x= ey.
Since∂M
∂y=∂N
∂x
so the given differential equation is exact. Now,∫eydx+
∫(0 + 2y)dy = c.
That implies
xey + y2 = c
is the desired solution.
Exercise: Solve the following differential equation
1. 3y2dx+ xdy = 0, y(1) = 12
2. 2 sin 2x sinh ydx− cos 2x cosh ydy = 0, y(0) = 1.
3. (3−y)x2
dx+ (y−2x)xy2
dy = 0, y(−1) = 2.
4. 2xydy = (x2 + y2)dx, y(1) = 2.
8 Integrating Factors
Definition 8.1. An integrating factor is a function when multiplied by it, the
left hand side of the equation (6) becomes an exact differential equation.
How to find integrating factor:
Theorem 8.2. If∂M∂y− ∂N
∂x
N= f(x),
a function of x alone, then e∫f(x)dx is an integrating factor of the equation (6).
Theorem 8.3. If∂M∂y− ∂N
∂x
M= f(y),
a function of y alone, then e−∫f(y)dy is an integrating factor of the equation (6).
19
Example 8.4.
Find an integrating factor and solve the initial value problem:
2 sin y2dx+ xy cos y2dy = 0, y(2) =
√π
2.
Here
Solution 8.5. Here M(x, y) = 2 sin y2 and N(x, y) = xy cos y2. Now
∂M
∂y= 4y cos y2 6= y cos y2 =
∂N
∂x.
Hence the differential equation is not exact. Now
∂M∂y− ∂N
∂x
N=
3
x,
Therefore the integrating factor is
e∫
3xdx = x3.
Multiplying the given equation by x3, we get the new equation
2x3 sin y2dx+ x4y cos y2dy = 0.
This equation is exact because
∂
∂y(2x3 sin y2) = 4x3y cos y2 =
∂
∂x(x4y cos y2).
Thus Solution of the given differential equation is
x4 sin y2
2= c = constant.
Substituting the initial condition y(2) =√
π2
in to the solution we have,
c = 8.
Hence the desired particular solution is
x4 sin y2 = 16.
Exercise: Find an integrating factor and solve the following differential equation
1. 2xydx+ 3x2dy = 0.
2. (2 cos y + 4x2)dx = x sin ydy.
3. x−1 cosh ydx+ sinh ydy = 0.
4. (xy − 1)dx+ (x2 − xy)dy = 0.
5. (sinx+ cosx tan y)(dx+ dy) + 2 sin ydy = 0.
20
9 Linear Differential equation, Bernoulli Equa-
tion
The differential equation of the form
y′ + p(x)y = r(x)
is called linear differential equation equation where as equation of the form
y′ + p(x)y = r(x)yn
is called Bernoulli Equation.
How to solve
Given equation of the form y′ + p(x)y = r(x).
Integrating factor µ = e∫P (x)dx.
Solution y = 1µ{∫r(x)µdx.}
Example 9.1.
y′ − y = e2x.
Solution 9.2. Here p = −1, r = e2x, µ = e∫(−1)dx = e−x.
y(x) = ex[∫e−xe2xdx+ c
]= cex + e2x
Solution of Bernoulli Equation
Example 9.3.
y′ − ay = −by2
Solution 9.4. Dividing y2 throughout the equation and taking u = y−1 we get
the resulting equation
u′ + au = b
which is a linear equation, taking p = a, r = b, we get µ = eax
The solution u = e−ax[baeax + c
]= ce−ax + b
a.
y = 1u
= 1( ba)+ce−ax
Home work
Solve the differential equation
21
1. y′ + 4y = cosx.
2. x2y′ + 2xy = sinh5x.
3. y′ = (y − 2) cotx.
4. y′ + xy = xy−1.
5. y′ = 16ey−2x
10 Linear Differential Equations of second and
higher order
A second order differential equation is called linear if it can be written
y′′ + p(x)y′ + q(x)y = r(x)
and nonlinear if it cannot be written in this form. If r(x) = 0 in the above
equation, then the differential equation is called homogeneous, otherwise non-
homogeneous.
Theorem 10.1. For a homogeneous linear second order differential equation,
any linear combination of two solutions on an open interval I is again a solution
of that equation on I. In particular, for such an equation, sums and constant
multiples of solutions are again solutions.
Proof. Let y1 and y2 be two solutions. The by substituting y = c1y1 + c2y2 and
its derivative into the homogeneous second order differential equation, we get
y′′ + py′ + qy = (c1y1 + c2y2)′′
+ p(c1y1 + c2y2)′+ q(c1y1 + c2y2)
= c1y′′
1 + c2y′′
2 + p(c1y′1 + c2y
′2) + q(c1y1 + c2y2)
= c1(y′′
1 + py′1 + qy) + c2(y′′
2 + py′2 + qy2)
= 0
This proves that y is a solution of the second order differential equation on I.
22
11 Second order Homogeneous Equations with
constant coefficients
Consider the second order homogeneous linear differential equation
y′′
+ ay′ + by = 0 (7)
whose coefficients a, b are constants. To Find the solution of the equation (7), we
put y = eλx, y′ = λeλx, y′′
= λ2eλx into the equation (7), we obtain
λ2 + aλ+ b = 0. (8)
This equation is called the characteristic equation (or auxiliary equation) of
(7). If the roots of the equation (8) are real and distinct (say λ1, λ2) then the
solution of the equation (7) is
y = c1eλ1x + c2e
λ2x.
If the roots of equation (8) are repeated real roots (say λ1 = λ2) then solution of
equation (7) is
y = (c1 + c2x)eλ1x.
In case of complex roots (say α± iβ) of equation (8) then the solution is
y = eαx(A cos βx+B sin βx)
where A,B are constants.
Example 11.1.
Solve y′′ − y′ − 6 = 0.
Solution 11.2. The auxiliary equation is λ2−λ− 6 = 0. That implies λ = 3,−2
which are real and distinct. Hence the general solution is
y = c1e3x + c2e
−2x.
Example 11.3.
Solve y′′
+ 8y′ + 16y = 0.
Solution 11.4. The characteristic equation is λ2 +8λ+16 = 0. It has the double
root λ = −4. Hence the general solution is
y = (c1 + c2x)e−4x.
23
Example 11.5.
Solve y′′
+ 4y′ + 5y = 0.
Solution 11.6. The auxiliary equation is λ2 + 4λ+ 5 = 0. Solving for λ, we get
λ = −2± i. Hence the general solution is
y = e−2x(A cosx+B sinx).
Home work
1. 10y′′
+ 6y′ − 4y = 0.
2. 9y′′ − 30y′ + 25y = 0.
3. y′′
+ 4y′ + 4y = 0.
4. y′′ − 5y′ + 6y = 0.
5. y′′
+ 0.2y′ + 4.01y = 0.
12 Euler-Cauchy Equation
Consider the equation
x2y′′
+ axy′ + by = 0, a, b constants (9)
which is called the Euler-Cauchy Equation. Working rule for finding the
Euler-Cauchy equation:
Put
y = xm, y′ = mxm−1, y′′ = m(m− 1)xm−2
into the (9) and equating the coefficients of xm to zero we obtain
m2 + (a− 1)m+ b = 0
which as called the auxiliary equation. Three cases of solution:
case1 If the auxiliary equation has distinct real roots m1, m2 the corresponding
general solution of (9) is
y = c1xm1 + c2x
m2 .
case2 If the auxiliary equation has double root say m1 = m2 the the corre-
sponding general solution is
y = (c1 + c2 lnx)xm1 .24
case3 If the auxiliary equation has complex conjugate roots, say, m1 = µ +
iν, m2 = µ− iν then the corresponding general solution of (9) is
y = xµ[A cos(ν lnx) +B sin(ν lnx)].
Example 12.1.
Solve the Euler-Cauchy Equation
x2y′′ − 2.5xy′ − 2.0y = 0
Solution 12.2. The auxiliary equation is m2 − 3.1m − 2 = 0. The roots are
m1 = −0.5, m2 = 4. So the general solution is y = c1x−0.5 + c2x
4.
Example 12.3.
Solve the Euler-Cauchy Equation
x2y′′ − 3xy′ + 4y = 0
Solution 12.4. The auxiliary equation has the double root m = 2. So the general
solution is y = (c1 + c2 lnx)x2.
Example 12.5.
Solve the Euler-Cauchy Equation
x2y′′ + 7xy′ + 13y = 0
Solution 12.6. The auxiliary equation is m2 + 6m + 13 = 0. The roots are
m1 = −3± 2i So the general equation is y = x−3(A cos 2 lnx+B sin 2 ln x).
Home work Solve
1. x2y′′ + xy′ + y = 0.
2. 10x2y′′ + 46xy′ + 32.4y = 0.
3. x2y′′ − 20y = 0.
4. 4x2y′′ + 24xy′ + 25y = 0.
5. x2y′′ − 2xy′ + 2y = 0.
13 Solution by undetermined coefficients
Consider the differential equation
y′′ + ay′ + by = r(x) (10)25
Table 1: Method of undetermined coefficients.
Term in r(x) Choice for yp
keγ(x) ceγx
kxn, n = 0, 1, 2, 3, · · · knxn + kn−1x
n−1 + · · · k1x+ k0
k cosωx k cosωx+m sinωx
k sinωx k cosωx+m sinωx
keα(x) cosωx eax(k cosωx+m sinωx)
keα(x) sinωx eax(k cosωx+m sinωx)
A general solution of a nonhomogeneous linear differential equation (10) is a sum
of of the form
y = yh + yp
where yh is a general solution of the corresponding homogeneous equation and yp
is any particular solution of nonhomogeneous equation. Here our main task is to
discuss methods for finding such yp.
Rules for the method od undetermined coefficients
(A) Basic Rule:If (r(x) in (10) is one of the functions in the first column
in the Table1, chose the corresponding function yp in the second column and
determined its undetermined coefficients by substituting yp and its derivatives
into (10).
Modification Rule:
If a term in your choice for for yp happens to be a solution of the homogenous
equation corresponding to (10), then multiply your choice of yp by x (or by x2
of this solution corresponds to a double root of the characteristic equation of the
homogeneous equation).
Sum rule
If r(x) is a sum functions in several lines of table 1, first column then choose
for yp the sum of the functions in the corresponding lines of the second column.
Example 13.1.
26
Solve the nonhomogeneous equation
y′′
+ 4y = 8x2.
Solution 13.2. By using the table (1), we choose
yp = K2x2 +K1x+K0.
Then y′′p = 2K2. Substitution gives
2K2 + 4(K2x2 +K1x+K0) = 8x2.
Equating the coefficients of x2, x, x0 on both sides, we have 4K2 = 8, 4K1 =
0, 2K2 + 4K0 = 0. Thus K2 = 2, K1 = 0, K0 = −1. Hence yp = 2x2 − 1, and a
general solution of the equation is y = yh + yp = A cos 2x+B sin 2x+ 2x2 − 1.
Exercise
Find a general solution of the given differential equation using method of unde-
termined coefficients.
1. y′′
+ 10y′ + 25y = e−5x.
2. y′′ + 2y′ − 35y = 12e5x + 37 sin 5x.
3. y′′ − y′ − 34y = 21 sinh2x.
4. y′′ + y′ + 9.25y = 9.25(4 + e−x).
5. y′′ + 1.2y′ + 0.36y = 4e−0.6x.
6. y′′ − 3y′ = 28 cosh4x.
7. 3y′′ + 10y′ + 3y = secx.
8. y′′ − 12y′ + y = −6x3 + 3x2 + 377 sinx.
14 Solution by variation of parameters
Consider the differential equation
y′′ + p(x)y′ + q(x)y = r(x) (11)
27
with arbitrary variable function p, q, and r that are continuous on some interval
I. This method gives a particular solution yp of (11) on I in the form
yp = −y1∫y2r
Wdx+ y2
∫y1r
Wdx
where y1, y2 form a basis of solutions of the homogeneous equation
y′′ + p(x)y′ + q(x)y = 0
corresponding to (11) and
W = y1y′2 − y2y′1
is the wronskian of y1, y2.
Example 14.1.
Find the particular integral of
y′′ + y = cosec x
using method of variation of parameters.
Solution 14.2. The complementary function is
yh = c1 cosx+ c2 sinx.
Here y1 = cosx, y2 = sinx. The Wronskian W = 1. Hence yp = −x cosx +
sinx log(sinx).
Home Work
Solve the following differential equations using method of variation of parame-
ters.
1. y′′ + y = secx.
2. y′′ + 4y′ + 4y = 2e−2x
x2.
3. y′′ − 2y′ − 3y = 2ex − 10 sinx.
4. y′′ + y = x sinx.
5. y′′ + 2y′ + 4y = cos 4x.
6. y′′ + 9y = cosec x
28
UNIT-IV
15 Series solution of Differential Equations
To solve the homogeneous differential equations with constant coefficients we
have used some algebraic methods. But to find or solve the differential equation
with variable coefficient such as Legender’s equations, Bessel’s equation we need
to use power series method.
i.e., Let
y′′ + p(x)y′ + q(x)y = 0,
we assume here a solution in the form of a power series with unknown coefficient
as,
y =∞∑n=0
anxn = a0 + a1x+ a2x
2 + · · ·
is a solution of a given differential equation.
Differentiating with respect to x
y′ =∞∑n=1
nanxn−1 = a1 + 2a2x+ 3a3x
2 + · · ·
and
y′′ =∞∑n=2
n(n− 1)anxn−2 = 2a2 + 6a3x+ · · · .
Substituting y, y′, y′′ in the given differential equation of the second order, we
have the power series form of differential equation.
To find the solution of differential equation, we have to get the values of
a0, a1, a2, · · · . To obtain these coefficients, we have to equate with the power of
x. After finding the values and putting in the given equation, we get the solution
of the given differential equation.
Example 15.1.
Solve
y′ = 3y. (12)
Solution 15.2. Let
y =∞∑n=0
anxn = a0 + a1x+ a2x
2 + · · ·
29
is a solution of a given differential equation.
Differentiating with respect to x, we have
y′ =∞∑n=1
nanxn−1 = a1 + 2a2x+ 3a3x
2 + · · · .
Substituting this in (12), we have
a1 + 2a2x+ 3a3x2 + · · · = 3a0 + 3a1x+ 3a2x
2 + · · · .
By comparing the coefficients of same power of x from above equation, we have
a1 = 3a0,
2a2 = 3a1 = 3(3a0) = 9a0,
⇒ a2 =9
2a0.
Again
3a2 = 3a3,
⇒ a3 = a2 =9
2a0.
Substituting these values in the solution, we have
y = a0 + 3a0x+9
2a0x
2 + · · · .
⇒ y = a0(1 + 3x+9
2x2 + · · · ).
Example 15.3.
Find the power series solution of
(1− x)y′ = y. (13)
Solution 15.4. Let
y =∞∑n=0
anxn = a0 + a1x+ a2x
2 + · · ·
is a solution of a given differential equation.
Differentiating with respect to x, we have
y′ =∞∑n=1
nanxn−1 = a1 + 2a2x+ 3a3x
2 + · · · .
Substituting this in (13), we have
(1− x)(a1 + 2a2x+ 3a3x2 + · · · ) = a0 + a1x+ a2x
2 + · · ·
30
⇒ a1 + 2a2x+ 3a3x2 + · · · − a1x− 2a2x
2 − 3a3x3 − · · · = a0 + a1x+ a2x
2 + · · · .
⇒ a1 + 2a2x+ 3a3x2 + · · · = a0 + 2a1x+ 3a2x
2 + · · · .
comparing the coefficients of x, we have
a0 = a1,
2a2 = 2a1 = a0,
⇒ a2 = a1 = a0.
3a3 = 3a2,
⇒ a3 = a2 = a1 = a0.
Substituting these values in the solution, we have
y = a0 + a0x+ a0x2 + · · · .
⇒ y = a0(1 + x+ x2 + · · · ).
Home Work
Solve the following differential equations using power series method.
1. y′ = 3x2y.
2. y′′ = y.
3. y′ + 2y = 0.
4. xy′ + 3y = 0.
16 Theory of Power-series Method; Radius of
Convergence
Consider the power series
∞∑n=0
an(x− x0)n = a0 + a1(x− x0) + a2(x− x0)2 + · · · , (14)
31
where a0, a1, a2, · · · are the constants called the coefficients of the power series.
x0 is a constant called the center of the series and x is a variable.
Some Expansion of Functions:
1
1− x= 1 + x+ x2 + · · · =
∞∑n=0
xn. (|x| < 1)
1
1 + x= 1− x+ x2 − x3 + · · · =
∞∑n=0
(−1)nxn.
ex = 1 + x+x2
2!+x3
3!+x4
4!+ · · · =
∞∑n=0
xn
n!.
cosx = 1− x2
2!+x4
4!− · · · =
∞∑n=0
(−1)nx2n
(2n)!.
sinx = x− x3
3!+x5
5!− · · · =
∞∑n=0
(−1)nx2n+1
(2n+ 1)!.
Let us take the nth partial sum of (14),
Sn(x) = a0 + a1(x− x0) + a2(x− x0)2 + · · ·+ an(x− x0)n.
If for some x = x1.
limn→∞
Sn(x1) = S(x1).
If S(x1) is finite, then the series convergent at x = x1, otherwise divergent.
17 Radius of Convergence or Circle of Conver-
gence:
A circle |x− x0| = R for which the power series of the form∑∞
n=0 an(x− x0)n
is convergent is known as circle of convergence and R is known as the radius of
convergence.
How to find the radius Of convergence of a power series:
The radius of convergence R of the power series can be calculated by
1
R= lim
n→∞|an+1
an|
32
or1
R= lim
n→∞sup |an|
1n .
Example 17.1.
Find the radius of convergence of the power series
∞∑m=0
x2m
m!.
Solution 17.2. Compare with the power series of the form∑∞
n=0 anxn.
Put x2 = z, Now the series becomes∑∞
m=0 amxm.
Here am = 1m!, and am+1 = 1
(m+1)!So
|am+1
am| = 1
(m+ 1)!.m! =
1
(m+ 1).
Now1
R= lim
m→∞
1
(m+ 1)= 0.
⇒ R =∞.
⇒ Radius of convergence is ∞.
Example 17.3.
Find the radius of convergence
∞∑n=1
(n+ 1)nxn,
and∞∑n=0
n!
nn(z + π)n.
Solution 17.4. In the first problem an = (n+ 1)n and an+1 = (n+ 2)(n+ 1).
1
R= lim
n→∞|an+1
an| = lim
n→∞|(n+ 2)(n+ 1)
(n+ 1)n| = lim
n→∞|(n+ 2)
n| = 1.
⇒ R = 1.
In the second problem an = n!nn , and an+1 = (n+1)!
(n+1)n+1 .
Now1
R= lim
n→∞|an+1
an| = lim
n→∞| (n+ 1)!
(n+ 1)n+1
nn
n!| = lim
n→∞| 1
(1 + 1n)n| = 1
e.
⇒ R = e.
Definition 17.5. A real function f(x) is called analytic at a point x = x0 if it can
be represented by a power series in powers of x − x0 with radius of convergence
R > 0.33
Home Work
Determine the radius of convergence of the following series.
1.∞∑m=0
(−1)mx4m.
2.∞∑m=0
mmxm.
3.∞∑m=0
x2m+1
(2m+ 1)!.
4.∞∑m=0
(x− x0)2m
2m.
5.∞∑m=0
(2
3
)mx2m.
18 Legendre’s Equation, Legendre Polynomials
Pn(x):
Consider
(1− x2)y′′ − 2xy′ + n(n+ 1)y = 0 (15)
where n is a real number and the Solution of (15) is Legendre function.
Dividing the coefficient of y′′, we have
y′′ − 2x
(1− x2)y′ +
n(n+ 1)
(1− x2)y = 0.
Now applying power series method; we have
y =∞∑m=0
amxm,
y′ =∞∑m=1
mamxm−1,
and
y′′ =∞∑m=2
m(m− 1)amxm−2.
34
Substituting the values in (15), we have
(1− x2)∞∑m=2
m(m− 1)amxm−2 − 2x
∞∑m=1
mamxm−1 + n(n+ 1)
∞∑m=0
amxm = 0.
By writing the first expression as two separate series, we have∞∑m=2
m(m−1)amxm−2−
∞∑m=2
m(m−1)amxm−2
∞∑m=1
mamxm+n(n+1)
∞∑m=0
amxm = 0.
By writing out each series and arranging each power in a column, we obtain,
2a2 + 6a3x+ 12a4x2 + · · ·+ (s+ 2)(s+ 1)as+2x
s · · ·
−2a2x2 − · · ·+ · · ·
−2a1x− 4a2x2 − · · · − s(s− 1)asx
s − · · ·
n(n+ 1)a0 + n(n+ 1)a1x+ n(n+ 1)a2x2 + · · · − 2asx
s − · · · = 0.
Equating the coefficients of each power of x to zero, we obtain,
n(n+ 1)a0 + 2a2 = 0.
[n(n+ 1)− 2]a1 + 6a3 = 0.
[n(n+ 1)− 5]a2 + 12a4 = 0.
and in generally for s = 2, 3, 4, · · · ,
as+2 = −(n− s)(n+ s+ 1)
(s+ 2)(s+ 1)as (s = 0, 1, 2, · · · )
This is called a recurrence relation. By inserting these values, we obtain
y(x) = a0y1(x) + a1y2(x)
where
y1(x) = 1− n(n+ 1)
2x2 +
(n− 2)n(n+ 1)(n+ 3)
4!x4 − · · ·
and
y2(x) = x− (n− 1)(n+ 2)
3!x2 + · · ·
The solution of Legendre’s differential equation is called the Legendre Polynomial
of degree n and is denoted by Pn(x). In general
Pn(x) =M∑m=0
(−1)m(2n− 2m)!
2nm!(n−m)!(n− 2m)!xn−2m
where M = n2
or n−12
whichever is an integer. The First few of these functions
are
P0(x) = 1
P1(x) = x
P2(x) =1
2(3x2 − 1)
P3(x) =1
2(5x3 − 3x)
35
19 Frobenious Method:
Consider
y′′ + P (x)y′ +Q(x)y = 0 (16)
This method will applied only for regular singular point if (16).
Singular Point:
A point x = x0 will be a singular point of (16), if P (x), Q(x) are not analytic
at x = x0.
Regular Singular Point:
Let x = x0 be a singular point of (16). Then the singular point is said to be
regular singular point if (x− x0)P (x) and (x− x0)2Q(x) are analytic at x = x0.
Example 19.1.
(x− 1)2y′′ + 2x(x− 1)y′ + 3(x+ 1)y = 0
⇒ y′′ +2x(x− 1)
(x− 1)2y′ +
3(x+ 1)
(x− 1)2y = 0.
Here P (x) =2x(x− 1)
(x− 1)2, Q(x) =
3(x+ 1)
(x− 1)2.
Clearly at x = 1, P (x) and Q(x) are not analytic. Now
(x− x0)P (x) = (x− 1)2x(x− 1)
(x− 1)2= 2x,
and
(x− x0)2Q(x) = (x− 1)23(x+ 1)
(x− 1)2= 3(x+ 1).
But both are analytic at x = 1. Now let x = x0 be a regular singular point of
(16), then the series solution of (16) can be assumed as y(x) =∞∑n=0
an(x−x0)n+m,
where m is real number satisfying the indicial equation which is of the type
m(m− 1) + p0m+ q0 = 0
where p0, q0 are constant terms of (x− x0)P (x) and (x− x0)2Q(x) respectively.
Case-I
Let m1, m2 be two distinct roots of the indicial equation, then the two Linearly
Independent solutions of (16) is given by
y1(x) =∞∑n=0
an(x− x0)n+m1
36
y2(x) =∞∑n=0
an(x− x0)n+m2 .
Case-II
Let m1 = m2 = m be double roots of the indicial equation, then a basis is
y1(x) = xm(a0 + a1x+ a2x2 + · · · )
y2(x) = y1(x) lnx+ xm(a1x+ a2x2 + · · · )
Case-II
Roots differ by an integer
A basis is
y1(x) = xm1(a0 + a1x+ a2x2 + · · · )
and
y2(x) = ky1(x) lnx+ xm2(a0 + a1x+ a2x2 + · · · )
where the roots are so denoted that m1−m2 > 0 and k may turn out to be zero.
Home work
Find a basis of solutions of the following differential equations.
1. xy′′ + 2y′ + 4xy = 0.
2. (x+ 1)2y′′ + (x+ 1)y′ − y = 0.
3. 2x(x− 1)y′′ − (x+ 1)y′ + y = 0.
4. x2y′′ + x3y′ + (x2 − 2)y = 0.
20 Bessel’s Equation, Bessel’s functions
Consider the Bessel’s differential equation
x2y′′ + xy′ + (x2 − ν2)y = 0
where ν ia any real non-negative number.Bessel’s equation can be solved by using
Frobenius method. Substituting y(x) =∑∞
m=0 amxm+r, y′, y′′ into the Bessel’s
equation and equating the coefficient of least power of x to zero, we obtain the
indicial equation
(r + ν)(r − ν) = 0.
37
Thus the solution of the Bessel’s differential equation is
Jn(x) = xn∞∑m=0
(−1)mx2m
22m+nm!(n+m)!.
This is called the Bessel function of the first kind of order n.
Gamma Function
The gamma function Γ(ν) defined by the integral
Γ(ν) =
∫ ∞0
e−ttν−1dt (ν > 0).
This yields a basic relationship
Γ(ν + 1) = νΓ(ν).
In general
Γ(n+ 1) = n! n = 0, 1, 2, · · ·
We list here some important recurrence relationships of Bessel’s functions:
d
dx[xνJν(x)] = xνJν−1(x)
d
dx[x−νJν(x)] = −x−νJν+1(x)
Jν−1(x) + Jν+1(x) =2ν
xJν(x).
Jν−1(x)− Jν+1(x) = 2J ′ν(x).
Γ(1
2) =√π.
J 12(x) =
√2
πxsinx.
J−12
(x) =
√2
πxcosx.
References
[1] Erwin Kreyszig, Advanced Engineering Mathematics, John Wiley and Sons,
Inc, New York, 2006.
[2] J. Sinha Roy and S.Padhy, A course on Ordinary and partial differential equa-
tions, Kalyani Publishers, 2010.
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